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2 years ago

Witch answer represents a speed, of velocity A-35m/s B-35 m/s south C- 35 m/s east -35 m/a down

Physics

1 answer:

sladkih [1.3K]2 years ago

3 0

Answer:

A 35 m/s

Explanation:

Speed of any object is scalar quantity while velocity is a vector. The scalar quantity is represented only by magnitude it doesn't need direction.

Hence, option A is correct as all other options have direction as well.

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A galvanometer can be calibrated to read electric A) current B) Voltage C) either of these D) neither of these

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3 years ago

The si unit of mass is the Newton. True or false.

LUCKY_DIMON [66]

False it’s the unit of force

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3 years ago

1. List one way mitosis and meiosis are the same and one way they are different.

kolbaska11 [484]

1. They both are same in the process, I mean they both are types of cell division they both form and give rise to new cells from the existing ones
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6 0

3 years ago

a lorry travels 3600m on a test track accelerating constantly at 3m/s squared from standstill. what is the final velocity (3 sig

suter [353]

The final velocity of the truck is found as 146.969 m/s.

Explanation:

As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.

Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.

$2as =v^{2} - u^{2} \\\\v^{2} = 2as + u^{2}$

Then, the final velocity will be

$v^{2} = 2 * 3 * 3600 + 0 = 21600\\ \\v=\sqrt{21600}=146.969 m/s$

Thus, the final velocity of the truck is found as 146.969 m/s.

8 0

3 years ago

How much heat is released when 432 g of water cools down from 71'c to 18'c?

maria [59]

The heat released by the water when it cools down by a temperature difference $\Delta T$ is
$Q=mC_s \Delta T$
where
m=432 g is the mass of the water
$C_s = 4.18 J/g^{\circ}C$ is the specific heat capacity of water
$\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ}$ is the decrease of temperature of the water

Plugging the numbers into the equation, we find
$Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J$
and this is the amount of heat released by the water.

7 0

3 years ago